Problem: $2bc + bd - b + 3 = -4c - 4$ Solve for $b$.
Explanation: Combine constant terms on the right. $2bc + bd - b + {3} = -4c - {4}$ $2bc + bd - b = -4c - {7}$ Notice that all the terms on the left-hand side of the equation have $b$ in them. $2{b}c + 1{b}d - 1{b} = -4c - 7$ Factor out the $b$ ${b} \cdot \left( 2c + d - 1 \right) = -4c - 7$ Isolate the $b$ $b \cdot \left( {2c + d - 1} \right) = -4c - 7$ $b = \dfrac{ -4c - 7 }{ {2c + d - 1} }$ We can simplify this by multiplying the top and bottom by $-1$. $b= \dfrac{4c + 7}{-2c - d + 1}$